Question

1如图,过平行四边形ABCD的顶点A作一条直线,分别交BD,CD及BC的延长线于GFE,若CE=1/2 BC,则DF/F

Answer 1

1.CEF与BEA相似,则AB／CF=BE／CE=3则AB=3CF,CD=3CF.所以F是CD三等分点,DF/FC =2

连接AC,G是ACD重心,则AG=2GF,CEF与BEA相似,AE／EF=BE／CE=3,则AE=3EF=AG+GF=3GF,则GF=EF,AG=2GF=GF+EF=GE,则AG/GE=1

2.CFG与EFD相似,则CG／DE=CF／EF=1／3,则DE=3CG,BDE与BAC相似,则DE／AC=BE／BC=3／5,则3AC=5DE,3（3CG）=（AG+CG）,可解得AG／CG=4